Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int grid[110][110], aux[110][110];
string p[2] = {"row", "col"};
void insere(vector<pair<int, int> > &v, int tipo, int qtd, int val) {
  if (qtd <= 0) return;
  while (qtd) {
    qtd--;
    v.emplace_back(tipo, val);
  }
}
int main(void) {
  int n, m;
  scanf("%d %d", &n, &m);
  bool flag = true;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &grid[i][j]);
      flag = flag and !grid[i][j];
    }
  }
  if (flag) {
    cout << 0 << endl;
    return 0;
  }
  vector<pair<int, int> > ans;
  for (int k = 0; k <= 500; k++) {
    vector<pair<int, int> > at;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) aux[i][j] = grid[i][j];
    bool flag = true;
    insere(at, 0, k, 0);
    for (int j = 0; j < m; j++) aux[0][j] -= k;
    for (int j = 0; j < m; j++)
      if (aux[0][j] > 0) {
        insere(at, 1, aux[0][j], j);
        for (int i = n - 1; i >= 0; i--) aux[i][j] -= aux[0][j];
      }
    for (int i = 1; i < n; i++) {
      insere(at, 0, aux[i][0], i);
      if (aux[i][0] < 0) flag = false;
      for (int j = m - 1; j >= 0; j--) aux[i][j] -= aux[i][0];
    }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (aux[i][j]) flag = false;
    if (flag and (ans.empty() or int(at.size()) < int(ans.size()))) ans = at;
  }
  if (ans.empty())
    printf("-1\n");
  else {
    cout << ans.size() << endl;
    for (auto v : ans) cout << p[v.first] << " " << v.second + 1 << endl;
  }
  return 0;
}